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3 years ago

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I need help plz plz plz

1 answer:

Katena32 [7]3 years ago

3 0

Your answer to this question is B

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Dwayne's garden is triangle-shaped with two equal sides and a third side that is 4 ft more than the length of an equal side. The

poizon [28]

D because its a triangle so it would be 3x but then you have to add the 4 for the longer side.

8 0

3 years ago

How tall would one million pennies be in centimeters.And how do you know

Vikki [24]

One US-penny is 19 mm in diameter.

One British-penny is 20.3 in diameter.

We can make an accurate measurement of the diameters, or research on the internet.

1 milion US_pennies are

1,000,000*19 mm = 19,000,000 mm = 1,900,000 cm tall, since 1cm=10mm

1 milion British_pennies are

1,000,000*20.3 mm = 20,300,000 mm = 2,030,000 cm tall, since 1cm=10mm

8 0

3 years ago

H=-16t^2+96t+4<br>solve by factoring

zlopas [31]

Answer:

6 plus or minus square root of 37/2

Step-by-step explanation:

Is the quadratic formula allowed or no?

4 0

3 years ago

Read 2 more answers

A random variable X follows the uniform distribution with a lower limit of 670 and an upper limit of 750.a. Calculate the mean a

DENIUS [597]

You can compute both the mean and second moment directly using the density function; in this case, it's

$f_X(x)=\begin{cases}\frac1{750-670}=\frac1{80}&\text{for }670\le x\le750\\0&\text{otherwise}\end{cases}$

Then the mean (first moment) is

$E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\frac1{80}\int_{670}^{750}x\,\mathrm dx=710$

and the second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f_X(x)\,\mathrm dx=\frac1{80}\int_{670}^{750}x^2\,\mathrm dx=\frac{1,513,900}3$

The second moment is useful in finding the variance, which is given by

$V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2=\dfrac{1,513,900}3-710^2=\dfrac{1600}3$

You get the standard deviation by taking the square root of the variance, and so

$\sqrt{V[X]}=\sqrt{\dfrac{1600}3}\approx23.09$

8 0

3 years ago

Thirty percent of the fifth grade students in a large school district read below grade level. The distribution of sample proport

fredd [130]

Answer: 0.9088

Step-by-step explanation:

Given : $\mu=0.30$

$\sigma=0.045$

Let x be a random available that represents the proportion of students that reads below grade level .

Using $z=\dfrac{x-\mu}{\sigma$ , for x= 0.36 , we have

$z=\dfrac{0.36-0.30}{0.045}=1.33333$

Using standard normal z-value table,

P-value $= P($

$P(z$ [Rounded yo the nearest 4 decimal places.]

Hence, the probability that a second sample would be selected with a proportion less than 0.36 = 0.9088

7 0

3 years ago